64 G. PLARR ON THE SOLUTION OF THE EQUATION Vpgp =0. 
1 ope 2 1+z 
Now as: =—5 <2 0, we have 3<ia242<1- 
Therefore whatever be the sign of z, and consequently of h, and Q’, the absolute 
values of E which satisfy (¢) must be comprised between the absolute values 

of hz=h, and h,z pits = 5 , and the former value will be the greater 
one. In fact, the limits are 
1°) when /, > 0: ne E> ay 
(c 
i 2°) when h, <0: hee Ba, 
The conditions (a), (5), have to be satisfied even when E’* and Q’E, take 
their greatest value, namely h? and h?(—/,h;), both being positive. Namely, 
having in that case: 
Ee hiz" , QE=hiz(1+2) , 
the conditions (a), 0), in virtue of (50 ter); after suppressing the factors h?, it 
respectively become ; 
(a) l+eto—i2 as 
(0) (hepa 2 H(1+z) >0. 
1+ 

Introducing into (6) the quantity w=—,, =, and decomposing (w+ 17—3 Ww 
into the factors =(w—2)(w—3), and decomposing w—2, w—4 into factors 
depending on z, we get the conditions under the form: 
(a) Ey SG Se +2) >0, 
(0) (1—2)(5 +2)(V3+1—2)(V3-1+4%) >0, 
and by approximate numbers into : 
(a) - ..(1,3798 ..... —-2)-(0,5798 ... +2) > 0, 
() - (V2) (F + 2) (2,782. - — 2]0,732.. + 2] > 0. 
Now the limits of z give: 7 
Pree | 1o2 50, : ee Us 
and the other factors likewise are positive in virtue of those limits. It follows 
that when condition (c) is satisfied by a certain value of E, the conditions (a), 
(2), will be fulfilled necessarily also, and the equation 
Fé) = 0 
